Question

A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizontally enters uniform, vertical 0.0800 T magnetic field and follows a circular arc of radius of 38.5 cm. What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuterium?

Answer 1

_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:

$r=\frac{mv}{qB}$

Where:

$\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}$

We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:

$\frac{r_d}{m_d}=\frac{v}{q_B_}$

Now, we multiply both sides by the magnetic field "B":

$\frac{Br_d}{m_d}=\frac{v}{q}$

Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton:

$r_p=\frac{m_pv}{qB}$

And substitute the quotient between velocity and charge:

$r_p=\frac{m_p}{B}(\frac{Br_d}{m_d})$

Now, we cancel out the magnetic field:

$r_p=\frac{m_pr_d}{m_d}$

Now, we substitute the values:

$r_p=\frac{(1.67\times10^{-27}kg)(0.385m)}{(3.34\times10^{-27}kg)}$

Solving the operations:

$r_p=0.193m=19.3cm$

Therefore, the radius is 19.3 cm.